Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long int powmod(long long int a, long long int b) {
  if (b == 0) return 1;
  long long int x = powmod(a, b / 2);
  long long int y = (x * x) % 1000000007;
  if (b % 2) return (a * y) % 1000000007;
  return y % 1000000007;
}
int level[100010];
int ptr;
int subsize[100010];
int dp[100010][23];
vector<int> g[100010], c[100010];
set<int> ans;
vector<int> tree[4 * 100010];
int basearray[100010];
int chainindex[100010];
int chainhead[100010], chainno, posinbase[100010];
void build(int node, int start, int end) {
  if (start == end) {
    for (int i = 0; i < (int)(c[basearray[start]].size()); i++)
      tree[node].push_back(c[basearray[start]][i]);
    sort((tree[node]).begin(), (tree[node]).end());
    if ((int)(tree[node].size()) > 10) tree[node].resize(10);
    return;
  }
  int mid = (start + end) / 2;
  build(2 * node, start, mid);
  build(2 * node + 1, mid + 1, end);
  for (int i = 0; i < (int)(tree[2 * node].size()); i++)
    tree[node].push_back(tree[2 * node][i]);
  for (int i = 0; i < (int)(tree[2 * node + 1].size()); i++)
    tree[node].push_back(tree[2 * node + 1][i]);
  sort((tree[node]).begin(), (tree[node]).end());
  if ((int)(tree[node].size()) > 10) tree[node].resize(10);
}
void query_tree(int node, int start, int end, int l, int r) {
  if (start > r || end < l) return;
  if (start >= l && end <= r) {
    for (int i = 0; i < (int)(tree[node].size()); i++)
      ans.insert(tree[node][i]);
    return;
  }
  int mid = (start + end) / 2;
  query_tree(2 * node, start, mid, l, r);
  query_tree(2 * node + 1, mid + 1, end, l, r);
}
int lca(int p, int q) {
  if (level[p] < level[q]) swap(p, q);
  int lg;
  for (lg = 1; (1 << lg) <= level[p]; lg++)
    ;
  lg--;
  for (int i = lg; i >= 0; i--) {
    if (level[p] - (1 << i) >= level[q]) p = dp[p][i];
  }
  if (p == q) return p;
  for (int i = lg; i >= 0; i--) {
    if (dp[p][i] != -1 && dp[p][i] != dp[q][i]) {
      p = dp[p][i];
      q = dp[q][i];
    }
  }
  return dp[p][0];
}
void dfs(int u, int d, int par) {
  level[u] = d;
  subsize[u] = 1;
  dp[u][0] = par;
  for (int v : g[u]) {
    if (v != par) {
      dfs(v, d + 1, u);
      subsize[u] += subsize[v];
    }
  }
}
void HLD(int curr, int prev) {
  if (chainhead[chainno] == -1) chainhead[chainno] = curr;
  chainindex[curr] = chainno;
  posinbase[curr] = ptr;
  basearray[ptr] = curr;
  ptr++;
  int sc = -1;
  for (int v : g[curr]) {
    if (v != prev) {
      if (sc == -1 || subsize[v] > subsize[sc]) sc = v;
    }
  }
  if (sc != -1) HLD(sc, curr);
  for (int v : g[curr]) {
    if (v != prev) {
      if (sc != v) {
        chainno++;
        HLD(v, curr);
      }
    }
  }
}
void query_up(int u, int v) {
  if (level[u] < level[v]) swap(u, v);
  while (chainindex[u] != chainindex[v]) {
    query_tree(1, 0, ptr - 1, posinbase[chainhead[chainindex[u]]],
               posinbase[u]);
    u = dp[chainhead[chainindex[u]]][0];
  }
  query_tree(1, 0, ptr - 1, posinbase[v], posinbase[u]);
  return;
}
void query(int u, int v, int a) {
  int l = lca(u, v);
  query_up(u, l);
  query_up(v, l);
  printf("%d ", min((int)(ans.size()), a));
  int tot = min((int)(ans.size()), a);
  for (int it : ans) {
    if (tot == 0) break;
    tot--;
    printf("%d ", it + 1);
  }
  printf("\n");
  ans.clear();
}
int main() {
  int n, m, q;
  scanf("%d", &n), scanf("%d", &m), scanf("%d", &q);
  memset(dp, -1, sizeof(dp));
  memset(chainhead, -1, sizeof(chainhead));
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d", &u);
    scanf("%d", &v);
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 0; i < m; i++) {
    int x;
    scanf("%d", &x);
    x--;
    c[x].push_back(i);
  }
  dfs(0, 0, -1);
  for (int j = 1; j < 21; j++) {
    for (int i = 0; i < n; i++) {
      if (dp[i][j - 1] != -1) dp[i][j] = dp[dp[i][j - 1]][j - 1];
    }
  }
  chainno = 0;
  HLD(0, -1);
  build(1, 0, ptr - 1);
  while (q--) {
    int u, v, a;
    scanf("%d", &u), scanf("%d", &v), scanf("%d", &a);
    u--, v--;
    query(u, v, a);
  }
  return 0;
}